All categories

3 years ago

Write the quadratic equation whose roots are -2 and 1, and whose leading coefficient is 4 .

Mathematics

1 answer:

Kaylis [27]3 years ago

7 0

Answer: the equation is

4x^2 + 4x - 12

Step-by-step explanation:

A quadratic equation is an equation in which the highest power of the unknown is 2.

The general form of a quadratic equation is expressed as

ax^2 + bx + c

Where

a is the leading coefficient

c is a constant

Assuming we want to write the quadratic equation in x, from the information given, the roots which are given are -2 and 1 and the leading coefficient is 4.

Therefore, the linear factors of the quadratic equation will be (x+2) and (x-1)

the equation becomes

(x+2)(x-1)

= x^2 - x +2x - 3

= x^2 + x - 3

Given a leading coefficient of 4, we will multiply the quadratic expression by 4. It becomes

4(x^2 + x - 3)

= 4x^2 + 4x - 12

You might be interested in

Mr. Miller has a 20-foot ladder, which he

dimulka [17.4K]

Answer:

8 ft

Step-by-step explanation:

the ladder will make a right angle triangle with the house. The hypotenuse of the triangle is the ladder, 20 ft, and the height of the triagnle is 18 ft.

Solve for the base of the right angle triangle using pythagorean's theorem:

c=20, a=18, b=?

$b^{2}=c^{2} - a^{2}\\b^{2}=20^{2} - 18^{2}$

$b=\sqrt{72}$

$b= 8.49 ft$

round to the nearest foot,

b = 8 ft

3 0

3 years ago

Afina-wow [57]

Both are right angel triangle.

8 0

4 years ago

Read 2 more answers

Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

worty [1.4K]

Answer:

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.